EDIT: My previous answer was completely wrong. Here's a corrected version.

First, note that this property descends to subgroups: every irrep of a subgroup is a summand of a restriction from the bigger group. Thus if all the irreps of the bigger group are 1-d, all the irreps of the small group will be their restrictions with the same one coming up multiple times.

If $G$ has $n$ 1-dimensional quaternionic representations whose sum is faithful, then $G$ must embed into the product of $n$ number of copies of the unit quaterions (that is, $SU(2)$). Consider the image of $G$ in any one of these representations, and then consider the image in $SO(3)$; since it is a finite subgroup of $SO(3)$, it is either cyclic, dihedral (including $C_2\times C_2$), or $A_4,S_4$ or $A_5$ (see Wikipedia). However, if it's nonabelian, this group has a real irrep (an irrep over the real numbers whose endomorphisms are $\mathbb{R}$, which is thus absolutely irreducible) which isn't 1-d, so it tensoring gives an irrep over the quaternions that isn't 1-d, so it must be cyclic or $C_2\times C_2$. This shows that the original subgroup of $SU(2)$ is either abelian, or the classic quaternion group.

Thus, $G$ is a subgroup of a product of cyclic and quaternion groups, and as we argued above, all such subgroups have this property.

EDIT: My previous answer was completely wrong. Here's a corrected version.

First, note that this property descends to subgroups: every irrep of a subgroup is a summand of a restriction from the bigger group. Thus if all the irreps of the bigger group are 1-d, all the irreps of the small group will be their restrictions with the same one coming up multiple times.

If $G$ has $n$ 1-dimensional quaternionic representations whose sum is faithful, then $G$ must embed into the product of $n$ number of copies of the unit quaterions (that is, $SU(2)$). Consider the image of $G$ in any one of these representations, and then consider the image in $SO(3)$; since it is a finite subgroup of $SO(3)$, it is either cyclic, dihedral (including $C_2\times C_2$), or $A_4,S_4$ or $A_5$ (see Wikipedia). However, if it's nonabelian, this group has a real irrep (an irrep over the real numbers whose endomorphisms are $\mathbb{R}$, which is thus absolutely irreducible) which isn't 1-d, so it tensoring gives an irrep over the quaternions that isn't 1-d, so it must be cyclic or $C_2\times C_2$. This shows that the original subgroup of $SU(2)$ is either abelian, or the classic quaternion group.

Thus, $G$ is a subgroup of a product of cyclic and quaternion groups, and as we argued above, all such subgroups have this property.

EDIT: It's also worth mentioning that this shows that the notion of a maximal quaternionic linear subgroup exists: it's the quotient by the intersection of the kernel of all maps to abelian or $Q_8$ groups.

I didn't quite finish a characterization, but certainly such a group must be a semidirect product of an abelian group with a product of $\mathbb{Z}/2\mathbb{Z}$'s.

If $G$ has $n$ 1-dimensional quaternionic representations whose sum is faithful, then $G$ must embed into the product of $n$ number of copies of the unit quaterions (that is, $SU(2)$). Consider the image of $G$ in any one of these representations; since it is a finite subgroup of $SU(2)$, it is either cyclic, dihedral, or a central extension of $A_3,A_4$ or $A_5$ (see Wikipedia). However, if it's one of the latter 3, this group has a different irrep over the quaternions that isn't 1-d, so it must be cyclic or dihedral.

Thus, $G$ is a subgroup of a product of cyclic and dihedral groups. The intersection of $G$ with the product of the normal cyclic subgroups inside the dihedrals (and all of the cyclic group factors) gives a normal subgroup of $G$, and the quotient of $G$ by this subgroup is a product of $\mathbb{Z}/2\mathbb{Z}$'s. Also, this quotient is split by just taking any preimage of the each of the generators of the product of $\mathbb{Z}/2\mathbb{Z}$'s. Thus, we have a semi-direct product.

I suspect you can use some character calculations to see that actually this is a product of cyclics and dihedrals, but it's not working out at the moment, and I wouldn't be totally shocked if there are some funny exceptions.

EDIT: My previous answer was completely wrong. Here's a corrected version.

First, note that this property descends to subgroups: every irrep of a subgroup is a summand of a restriction from the bigger group. Thus if all the irreps of the bigger group are 1-d, all the irreps of the small group will be their restrictions with the same one coming up multiple times.

If $G$ has $n$ 1-dimensional quaternionic representations whose sum is faithful, then $G$ must embed into the product of $n$ number of copies of the unit quaterions (that is, $SU(2)$). Consider the image of $G$ in any one of these representations, and then consider the image in $SO(3)$; since it is a finite subgroup of $SO(3)$, it is either cyclic, dihedral (including $C_2\times C_2$), or $A_4,S_4$ or $A_5$ (see Wikipedia). However, if it's nonabelian, this group has a real irrep (an irrep over the real numbers whose endomorphisms are $\mathbb{R}$, which is thus absolutely irreducible) which isn't 1-d, so it tensoring gives an irrep over the quaternions that isn't 1-d, so it must be cyclic or $C_2\times C_2$. This shows that the original subgroup of $SU(2)$ is either abelian, or the classic quaternion group.

Thus, $G$ is a subgroup of a product of cyclic and quaternion groups, and as we argued above, all such subgroups have this property.